Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

13.3: Changing to a Basis of Eigenvectors

If we are changing to a basis of eigenvectors, then there are various simplifications:

1.    Since \(L:V\to V\), most likely you already know the matrix \(M\) of \(L\) using the same input basis as output basis \(S=(u_{1},\ldots ,u_{n})\) (say).

2.    In the new basis of eigenvectors \(S'(v_{1},\ldots,v_{n})\), the matrix \(D\) of \(L\) is diagonal because \(Lv_{i}=\lambda_{i} v_{i}\) and so
$$
\big(L(v_{1}),L(v_{2}),\ldots,L(v_{n})\big)=(v_{1},v_{2},\ldots, v_{n})
\begin{pmatrix}
\lambda_{1}&0&\cdots&0\\
0&\lambda_{2}&&0\\
\vdots&&\ddots&\vdots \\
0&0&\cdots&\lambda_{n}\end{pmatrix}\, .
$$

3.    If \(P\) is the change of basis matrix from \(S\) to \(S'\), the diagonal matrix of eigenvalues \(D\) and the original matrix are related by \(D=P^{-1}MP\).

This motivates the following definition:

 

Definition

A matrix \(M\) is \(\textit{diagonalizable}\) if there exists an invertible matrix \(P\) and a diagonal matrix \(D\) such that

\[ D=P^{-1}MP. \]
 

We can summarize as follows:

  1. Change of basis rearranges the components of a vector by the change of basis matrix \(P\), to give components in the new basis.
  2. To get the matrix of a linear transformation in the new basis, we \(\textit{conjugate}\) the matrix of \(L\) by the change of basis matrix: \(M\mapsto P^{-1}MP\).
     

If for two matrices \(N\) and \(M\) there exists a matrix \(P\) such that \(M=P^{-1}NP\), then we say that \(M\) and \(N\) are \(\textit{similar}\).  Then the above discussion shows that diagonalizable matrices are similar to diagonal matrices.


Corollary
A square matrix \(M\) is diagonalizable if and only if there exists a basis of eigenvectors for \(M\).  Moreover, these eigenvectors are the columns of the change of basis matrix \(P\) which diagonalizes \(M\).

 

Example 122

Let's try to diagonalize the matrix

\[M=\begin{pmatrix}
-14 & -28 & -44 \\
-7 & -14 & -23 \\
9 & 18 & 29 \\
\end{pmatrix}.\]

The eigenvalues of \(M\) are determined by

\[\det(M-\lambda I)=-\lambda^{3}+\lambda^{2}+2\lambda=0.\]

So the eigenvalues of \(M\) are \(-1,0,\) and \(2\), and associated eigenvectors turn out to be 
$$
v_{1}=\begin{pmatrix}-8 \\ -1 \\ 3\end{pmatrix},~~ v_{2}=\begin{pmatrix}-2 \\ 1 \\ 0\end{pmatrix}, {\rm ~and~~} v_{3}=\begin{pmatrix}-1 \\ -1 \\ 1\end{pmatrix}.
$$ 
In order for \(M\) to be diagonalizable, we need the vectors \(v_{1}, v_{2}, v_{3}\) to be linearly independent. Notice that the matrix
\[P=\begin{pmatrix}v_{1} & v_{2} & v_{3}\end{pmatrix}=\begin{pmatrix}
-8 & -2 & -1 \\
-1 & 1 & -1 \\
3 & 0 & 1 \\
\end{pmatrix}\]
is invertible because its determinant is \(-1\). Therefore, the eigenvectors of \(M\) form a basis of \(\Re\), and so \(M\) is diagonalizable. 
Moreover, because the columns of \(P\) are the components of eigenvectors, 
$$
MP=\begin{pmatrix}Mv_{1} &Mv_{2}& Mv_{3}\end{pmatrix}=\begin{pmatrix}-1.v_{1}&0.v_{2}&2.v_{3}\end{pmatrix}=\begin{pmatrix}v_{1}& v_{2} & v_{3}\end{pmatrix}\begin{pmatrix}
-1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 2 \\
\end{pmatrix}\, .
$$
Hence, the matrix \(P\) of eigenvectors is a change of basis matrix that diagonalizes \(M\):
\[P^{-1}MP=\begin{pmatrix}
-1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 2 \\
\end{pmatrix}.\]

 

 

 

Contributor